For any real number x, let [x] denote the largest integer less than or equal to x. If $$ \mathrm{I}=\int_{0}^{10}\left[\sqrt{\frac{10 \mathrm{x}}{\mathrm{x}+1}}\right] \mathrm{dx}, $$ then the value of 9I is ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Definite Integrals

For any real number x, let [x] denote the largest integer less than or equal to x. If

I = \int_{0}^{10} [\frac{10 x}{x + 1}] dx,

then the value of 9I is ____.

Answer: 182

Let

f (x) = \frac{10 x}{x + 1}

. Then

f^{'} (x) = \frac{10}{2 ( x + 1 ) ^{2} \frac{10 x}{x + 1}} > 0

, so f is increasing on [0, 10].

We need to find where

f (x) = 0, 1, 2, 3

$f (x) = 0$ when $x = 0$
$f (x) = 1$ when $\frac{10 x}{x + 1} = 1$ , i.e., $10 x = x + 1$ , so $x = 1/9$
$f (x) = 2$ when $\frac{10 x}{x + 1} = 4$ , i.e., $10 x = 4 x + 4$ , so $x = 2/3$
$f (x) = 3$ when $\frac{10 x}{x + 1} = 9$ , i.e., $10 x = 9 x + 9$ , so $x = 9$
At $x = 10$ : $f (10) = \frac{100}{11} \approx 3.015$

Therefore:

I = \int_{0}^{1/9} 0 d x + \int_{1/9}^{2/3} 1 d x + \int_{2/3}^{9} 2 d x + \int_{9}^{10} 3 d x

= 0 + (2/3 - 1/9) + 2 (9 - 2/3) + 3 (10 - 9)

= 5/9 + 2 (25/3) + 3 = 5/9 + 50/3 + 3 = \frac{5 + 150 + 27}{9} = \frac{182}{9}

Therefore,

9 I = 182

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Integer answer